The earth has a net electric charge that causes a field at points near its surface equal to 150150150 N/CN/CN/C and directed in toward the center of the earth. What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100100 m? Is use of the earth's electric field a feasible means of flight? Why or why not? Note: Part (a) asked for what magnitude and sign of charge would a 606060-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field.

Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 28b

Chapter 21, Problem 28b

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N/C$ and directed in toward the center of the earth. What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of $100$ m? Is use of the earth's electric field a feasible means of flight? Why or why not? Note: Part (a) asked for what magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field.

Verified step by step guidance

First, let's understand the concept of electric field. The electric field (E) is a vector field around a charged object where a force would be exerted on other charged objects. It is defined as the force per unit charge:

E = \frac{F}{q}

, where F is the force and q is the charge.

In part (a), you would calculate the charge of a person using the electric field of the Earth. Given that the electric field is 150 N/C, you can rearrange the formula to find the charge:

q = \frac{F}{E}

. Assume a typical force exerted on a person due to Earth's electric field to find q.

Now, for part (b), we need to calculate the force of repulsion between two people each with the charge calculated in part (a). Use Coulomb's Law, which states that the force between two point charges is given by:

F = \frac{k q^{2}}{r^{2}}

, where k is Coulomb's constant (

8.99 \times 10^{9} \frac{N}{m^{2}}

), q is the charge, and r is the distance between the charges (100 m in this case).

Substitute the values into Coulomb's Law to find the force of repulsion. Remember that the charge q is the one calculated from part (a).

Finally, consider whether the Earth's electric field is a feasible means of flight. Discuss the magnitude of the force calculated and compare it to the gravitational force acting on a person. Consider practical aspects such as the strength of the electric field required to counteract gravity and the implications for human flight.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

An electric field is a region around a charged object where other charges experience a force. It is represented by the symbol E and measured in newtons per coulomb (N/C). The direction of the field is defined as the direction a positive test charge would move if placed in the field. In this scenario, the Earth's electric field is directed towards its center.

Coulomb's Law

Coulomb's Law describes the force between two point charges. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It is given by the formula F = k * (|q1 * q2| / r^2), where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

Feasibility of Electric Field for Flight

The feasibility of using the Earth's electric field for flight involves considering the magnitude of the force it can exert. Given the relatively weak electric field strength of 150 N/C, the force generated would be insufficient to counteract gravitational forces for human flight. Thus, relying on Earth's electric field for flight is not practical with current technology and understanding.

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Intro to Electric Fields

Related Practice

Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, $1.60$ cm distant from the first, in a time interval of $3.20 \times 10^{- 6}$ s. Find the magnitude of the electric field.

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Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, $1.60$ cm distant from the first, in a time interval of $3.20$\times$10^{-6}$ s. Find the speed of the proton when it strikes the negatively charged plate.

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Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. How much time does it take the proton to stop after entering the field?

Textbook Question

Calculate the magnitude and direction (relative to the $+ x$ -axis) of the electric field in Example $21.6$ . Example $21.6$ : A point charge $q = - 8.0$ nC is located at the origin. Find the electric-field vector at the field point $x equals 1.2$ m, $y = - 1.6$ m.

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Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N / C$ and directed in toward the center of the earth. What magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field?

Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? Note: Part (a) asks for how much time does it take the proton to stop after entering the field.

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